Magnetic Force - AP Physics 2

A charged particle moving through a perpendicular magnetic field feels a Lorentz force equal to the formula:

is the charge, is the particle speed, and is the magnetic field strength. This force is always directed perpendicular to the particle’s direction of travel at that moment, and thus acts as a centripetal force. This force is also given by the equation:

We can set these two equations equal to one another, allowing us to solve for the radius of the arc.

Once the particle travels through a semicircle, it is laterally one diameter in distance from where is started (i.e. twice the radius of the circle).

Twice this value is the lateral offset of its crash point from the entrance:

The equation for finding the force on a moving charged particle in a magnetic field is as follows:

Here, is the force in Newtons, is the charge in Coulombs, is the velocity in , and is the magnetic field strength in Teslas.

Another way to write the equation without the cross-product is as follows:

Here, is the angle between the particles velocity and the magnetic field.

For our problem, because theta is , evaluates to 1, so we just need to perform multiplication.

Therefore, the force on the particle is 0.3N.

The equation for the force on a current carrying wire in a magnetic field is as follows:

is the force in Newtons, is the current in amperes, is the magnetic field strength in Teslas, and is the angle from parallel to the magnetic field.

Because our wire is not fully perpendicular to the magnetic field, it does not experience the full possible force. Instead, it experiences times the maximum force value.

The equation for force on a charge moving through a magnetic field is:

Because the velocity is perpendicular to the field, the cross product doesn't matter, and we can do simple multiplication.

Therefore, the force on the charge is

The equation for force on a charge moving through a magnetic field is:

.

The cross product is:

Above, is the degree from parallel the charge is moving. The charge is moving at from parallel, so the equation, once we plug in our numbers, is:

The force on the charge is about .

The force experienced by a charged particle in a magnetic field is

This means that when a charge particle is moving perpendicular to the field, due to the cross-product, it experiences the most amount of force (because is equal to , and theta equals when it's perpendicular). This means that the charged particle will experience no force due to the magnetic field when it's parallel. We know there will be no force on the particle, and we also know that uncharged particles experience no force in magnetic fields, but we can't say for certain the particle has no net charge, only being told that it's moving parallel to the field.

To find the force experienced by the charge, we use this equation:

Because the charge is moving at an angle from perpendicular, we need to take the cross product into account.

Theta is the angle from perpendicular, which is . Plug in known values and solve.

The rod acts as a battery due to its motion in the magnetic field: . Because there is a closed circuit, this results in current flow:

In this simple circuit, the current is the same everywhere, so the same current flows through the rod. Because of this current, the rod feels a force:

By the right-hand rule, this magnetic force is directed to the left, so the external force must be directed to the right. It is interesting to note that the power dissipated in the resistor:

is the same as the power provided by the outside force:

The universe conspires to conserve energy.

This question is presenting us with a scenario in which a charged particle is traveling with a certain velocity through a magnetic field. In this situation, we're being asked to determine what the force experienced by this particle is.

To solve this question, we'll need to determine what kind of force this particle is likely to experience. Since we're told that the particle is traveling in a magnetic field, it would make sense that this particle is going to be affected by a magnetic force. Thus, we'll need to use the equation for magnetic force.

Moreover, since we're told in the question stem that this particle is traveling perpendicularly to the magnetic field, we know that and thus . This helps reduce the equation down.

Now, all we need to do is plug in the values given to us in order to calculate the resulting force.

Use the magnetic force equation:

Plug in known values and solve for

Using the right hand rule for a current carrying wire shows that the magnetic field is pointing out of the screen. Using the right hand rule for magnetic force on a negatively charged particle shows the force acting downward.

In this question, we're asked to determine which answer choice falsely represents a necessary condition for a magnetic force to act on a particle.

To answer this, it is useful to look at the equation for magnetic force.

What this equation shows is that the magnetic force on a particle is dependent upon that particle's charge, its velocity, and on the strength of the magnetic field. Already we can rule out a few of the answer choices.

Additionally, this equation also shows that cannot be equal to zero. What this means is that the particle's direction of motion cannot be along the magnetic field lines. In other words, the particle cannot be travelling in a direction that is parallel or antiparallel with the magnetic field.

Lastly, we can see that no where in the above equation is there a variable for the size of the particle. Thus, size is not a requirement for a particle to experience a magnetic force.

Using

Where

is the charge limited to traveling in a single dimension

is the free charge

is the distance between the charges

is the velocity of the first charge

is the velocity of the second charge

is the value of the first charge

is the value of the second charge

and are equivalent as the charges are running parallel to each other

Combining equations:

Solving for

Plugging in values:

Finding the magnetic field at the location of the proton.

Converting to and plugging in values

Using

The centripetal force will need to be equal to the magnetic force.

Solving for

Converting:

Plugging in values:

Since the electron is stationary, there will be no magnetic force, as magnetic force requires the particle to be both charged and to be moving.

Magnetic fields do not affect charges that are moving parallel to them.

Thus, the magnetic force will be at a maximum when moving perpendicular to the field, and at zero when moving parallel to the field.

A charged particle moving through a perpendicular magnetic field feels a Lorentz force equal to the formula:

is the charge, is the particle speed, and is the magnetic field strength. This force is always directed perpendicular to the particle’s direction of travel at that moment, and thus acts as a centripetal force. This force is also given by the equation:

We can set these two equations equal to one another, allowing us to solve for the radius of the arc.

Once the particle travels through a semicircle, it is laterally one diameter in distance from where is started (i.e. twice the radius of the circle).

Twice this value is the lateral offset of its crash point from the entrance:

The equation for finding the force on a moving charged particle in a magnetic field is as follows:

Here, is the force in Newtons, is the charge in Coulombs, is the velocity in , and is the magnetic field strength in Teslas.

Another way to write the equation without the cross-product is as follows:

Here, is the angle between the particles velocity and the magnetic field.

For our problem, because theta is , evaluates to 1, so we just need to perform multiplication.

Therefore, the force on the particle is 0.3N.